Solving $e^{4x}+3e^{2x}-28=0$

Question

How to solve this equation:

$$e^{4x}+3e^{2x}-28=0$$

I don't know how to solve this problem. I read over another example, $e^{2x}-2e^x-8=0,$ and it said that $e^{2x}$ is $e$ to the $x$ squared, so $e^{2x}=e^{{x}^{2}}$. Why is this so? So $e^{4x}=e^{{x}^{4}}$?

Please help me to solve it.

Answer 1

Hint:

$$\eqalign{e^{4x}+3e^{2x}-28=0&\implies (e^{2x})^2+3(e^{2x})-28=0\\&\implies e^{2x}=\ldots\\&\implies x\:\:\:=\ldots}$$

And I think what "$e$ to the $x$ squared" means is $(e^{x})^2$, in this case it is equal to $e^{2x}$ since one have $(a^{m})^n=a^{mn}.$

Answer 2

Well, $e^{2x}=(e^x)^2$, but neither is equal to the "$e^{x^2}$" that you mentioned. So no, $e^{4x} \not= e^{x^4}$, but rather $e^{4x}=(e^4)^x$.

Therefore, $e^{4x}+3e^{2x}-28=0$ means $(e^{2x})^2+3(e^{2x})-28=0$.

Let $m=e^{2x}$, and you well get an auxiliary trinomial of $$m^2+3m-28=0.$$ If you factor your auxiliary trinomial, you will have $$(m+7)(m-4)=0.$$ Given the roots of $m=-7$ and $m=4$, can you solve for $x$ from here?

Answer 3

$$(e^{2x})^2+3(e^{2x})-28=0.$$ $$e^{2x}=y$$ $$y^2+3y-28=0$$ $$y_{1,2}=\frac{-3\pm11}{2}$$ $$y_1=4=e^{2x},2x=\ln 4$$ $$x=\ln2$$ for $$y_2=-7$$ no real solutions

Answer 4

$$ \Big(e^{2x}\Big)^2 + 3e^{2x}-28=0 $$ $$ u^2 + 3u - 28=0 $$ This is a quadratic equation. Once you've found $u$, you've got $e^{2x}$. After that, find $x$.